A rod with circular cross-section of radius 5 mm is stretched such that its cross-section remains circular. If after stretching the rod its length becomes four times the original length, then what is the radius of the new cross-section ?<\/p>\n
[amp_mcq option1=”1.5 mm” option2=”2.0 mm” option3=”2.5 mm” option4=”3.0 mm” correct=”option3″]<\/p>\n
The rod is stretched such that its cross-section remains circular. This implies the stretched rod is still a cylinder.
\nAfter stretching, the new length becomes four times the original length: $L_2 = 4 L_1$.
\nLet the new radius of the circular cross-section be $r_2$.
\nThe new volume of the rod is $V_2 = \\pi r_2^2 L_2 = \\pi r_2^2 (4L_1)$.<\/p>\n
Assuming the material of the rod is incompressible (a common assumption in such problems unless otherwise stated, meaning the density remains constant), the volume of the rod remains constant during stretching.
\nTherefore, $V_1 = V_2$.
\n$25\\pi L_1 = \\pi r_2^2 (4L_1)$.<\/p>\n
We can cancel $\\pi$ and $L_1$ from both sides (assuming $L_1 > 0$):
\n$25 = 4 r_2^2$.<\/p>\n
Solve for $r_2^2$:
\n$r_2^2 = \\frac{25}{4}$.<\/p>\n
Solve for $r_2$:
\n$r_2 = \\sqrt{\\frac{25}{4}} = \\frac{\\sqrt{25}}{\\sqrt{4}} = \\frac{5}{2}$.
\n$r_2 = 2.5$ mm.<\/p>\n